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July 10, 2003
Probability & Psychology
Let's play a game. Feel free to answer in the comment box. If you're familiar with the problem, don't give it away, but if you figure it out, please do. Please don't Google it either, I'll put up the link tomorrow afternoon. The Monty Hall Problem On the old television game show Let's Make a Deal, there was a segment where the host, Monty Hall, would present a contestant with three doors. Hidden behind one of these doors would be a prize and if the contestant guessed the right door he won the prize. A fairly simple guess game with a 1 in 3 shot of payoff. Every now and again though, Monty would make things a little more interesting by giving the contestant an additional choice. After the contestant had picked one of the three doors, Monty would then open one of the other two unselected doors which he knew to be empty. He would then ask the contestant if they would like to remain with their first choice or switch over to the remaining unopened door. The question is  In such a scenario would a contestant increase their chances of winning if they a) stick with their original choice, b) switch to the remaining unopened door, or c) it doesn't make a bit of difference either way? Update with answer: Well, a few brave souls in the comment box answered with the intuitive yet wrong answer of 'it doesn't matter either way', knowing full well that something had to be up or the question probably wouldn't have been asked ("who was buried in Grant's tomb" trick trick questions not withstanding), but no one got obstinate, like I was really hoping for. The best way to illustrate this, as Sliggy mentioned in the box, is through extreme exaggeration. What if instead of three there were a billion doors you had to choose from, and the prize was behind only one? Your odds of picking the right door on one try are now 1 in a billion, which is to say zero. But now let's say after you pick Monty Hall goes through and opens all those millions of empty doors, except for one, and asks you if you want to switch. The only way for that other one not to be the right door is if you picked the 1 in a billion door on your first shot! . . . So still think it's 1 in 2? I came across this problem on the evo psych discussion group in an article linked here. What interested me is the puzzle, yes, but perhaps more so the way in which people are drawn towards intuitive answers (and how that knowledge might be useful), and how accounts of this puzzle, with amusing consistency, always seem to indicate stubborn and/or angry reactions to the information (google it and see what I mean). Take this from the article: Indeed, some individuals I have encountered are so convinced that their (faulty) reasoning is correct that when you try to explain where they are going wrong, they become passionate, sometimes angry, and occasionally even abusive. And, just b/c we're Gene Expression, I'll quote Charles Murray on it too: . . .you should know for future reference that, stuck at a boring dinner party, you can create instant chaos, and sometimes rupture long and close friendships, by introducing the Monty Hall problem. Update #2: Wait, I gave up too soon, it appears Burbridge refuses to submit!: " . . . but the probability is still equal for each door. By opening a door, he gives you information about where the prize is not, but he gives you no information about where among the remaining options it is. Therefore there is no advantage in switching from your original choice. The title of my post has been justified. Score!! :P
Posted by Jason Malloy at
12:59 AM
I *think* it won't make a difference. There are two doors remaining, one of which hides the prize. So the contestant has a 50% chance of opening the right one. Which one he chooses doesn't change the probabilities, does it? The odds change from oneinthree to oneintwo, but the identity of the door isn't given away. Always switch. You have a 66% chance of winning vs. 33%. Here's another one. You are given a choice of flying a twin engine airplane which needs one working engine to fly, or a three engine airplane which needs two working engines to fly. Any one engine has the same probability of failure in either plane. Which plane is safer? Ole Posted by: Ole Eichhorn at July 10, 2003 02:05 AMI agree with Chris. There is no advantage in switching. Of course, this assumes that the host is not being malicious. If he is, then he is actually trying to tempt you away from the door you chose originally, because he knows the prize is behind it! Posted by: David B at July 10, 2003 03:08 AM
The odds change because Monty doesn't pick his door randomly. He tells you something about where the prize is by not picking that door. Posted by: Otto Kerner at July 10, 2003 06:21 AMSeen this one before (I did get the correct answer at the time  11 years ago I guess). Ole's problem: the twoengine plane is safer. Both planes crash and burn if two engines fail, and getting two engine failures is obviously more likely when you've got three running instead of two. My contribution to the list of problems: Someone offers you two envelopes, each containing a check made out to you. You know that one check is three times the amount of the other. You are given one of the envelopes and allowed to open it and see the amount. Should you switch to the other envelope? How should you decide whether to switch? Posted by: bbartlog at July 10, 2003 07:19 AMThis one was on Brad Delong awhile back. (Also Vos Savant further back). Very few people, including people with econ backgrounds, get it right intuitively. I didn't. Even figuring it out after you know the right answer isn't that easy. One way to explain it is that your second choice is not the first move of a new twodoor game. It's the second move of the original threedoor game. When Monte opened the empty door, he gave you information that you wouldn't have in a new twodoor game. Originally you knew that either you had chosen the right door (one chance) or else not (two chances). When he opened the door, the chances did not change  still only one chance that you've chosen the right door. So now the two chances are with the door he didn't open. Thinking of it as a new twodoor game is sort of like assuming that the cards are shuffled after each hand in a card game, when they're not. This is sort of a countingcards game, but with only three cards it's so simple that people miss it. Ole: your problem would be less trivial if you changed "three engine airplane" to "four engine airplane." Posted by: louis at July 10, 2003 10:02 AMThe problem with introducing a fourengine plane is that you now need more information to provide a definite answer. Assuming that you're talking about a fourengine plane that can fly with (max) two engines out, versus a twoengine plane that can fly with one engine out, then the fourengine plane is mostly safer  unless the engine failure rate is greater than 50%, in which case the twoengine plane is safer after all (I think  haven't formally analyzed it). Posted by: bbartlog at July 10, 2003 10:40 AMWell, the probability of the twinengine plane going down is p^2 the 4engine plane will go down if all 4 engines fail, or if exactly 3 engines fail (which can happen in 4 ways). The total probability of a disaster is p^4 + (4 p^3 (1p)) Both probabilities are equal when p=1/3. Take the four engine plane unless p exceeds 1/3. I've found good gamblers get this right when first asked. If you can't figure it out simply write out all the posibilites for a scenario when the prize is behind a given door... Posted by: carter at July 10, 2003 11:32 AMThere are actually 3 cases to consider. Case 3 is usually the unstated assumption, which causes some confusion. 1. Monty loves you and wants you to win. In this case he will only offer you the switch if your first pick was wrong. Obviously you should switch. When Monty loves you, you get a 100% chance of winning by always switching when offered. 2. Monty hates you and wants you to lose. Now he will only offer you the switch if your first pick was right, hoping to throw you off. In this case, don't switch. When Monty hates you, switching reduces your chance of winning from 1/3 to zero. 3. Monty is indifferent, and his decision whether to offer you a swtich is independent of whether your first pick was correct. In this case, switching increases your chance of winning to 2/3. 4. More generally, say Monty offers you a switch with probability p on a correct first pick, but with probability q on an incorrect first pick. Then switching is the right thing to do iff the ratio p/q is less than 2. Posted by: Daniel Lam at July 10, 2003 12:11 PMThat's a good point. Jason has actually made the problem more complicated by having Monty sometimes offer you the chance to switch, sometimes not. As the problem was first presented to me, he would always offer you the opportunity to switch. Posted by: bbartlog at July 10, 2003 01:05 PMIf you don't know whether Monty loves, hates, or doesn't care about you, as a player you're still in the 2/3 game though. Your choice remains the same, though the odds on the outcome, based on information you don't have, changes. But there's a third twist. Monty doesn't know whether you understand the odds or not. I f you understand the odds, what he should do if he hates you is clear. Otherwise he can't help or hurt you. Posted by: zizka at July 10, 2003 01:21 PMThis always seemed counterintuitive to me until somebody suggested you pick one door out of a thousand and Monty opens 998 doors and then offers you the chance to switch. Of course you switch. Posted by: sliggy at July 10, 2003 02:31 PMHuh??? Assume there are 1000 doors. Assume that the host is neither helpful nor malicious. Assume that he always opens one or more doors, but never opens one with the prize behind it (of course he knows where the prize is). When you make your initial choice, there is an equal 1in1000 chance that the prize is behind any door. Every time the host opens a door, he reduces the number of doors to choose from. He therefore increases the probability that the prize is behind any given remaining door, but the probability is still equal for each door. By opening a door, he gives you information about where the prize is not, but he gives you no information about where among the remaining options it is. Therefore there is no advantage in switching from your original choice. If anyone disagrees with this, can I gamble with them? Posted by: David B at July 11, 2003 02:09 AMTherefore there is no advantage in switching from your original choice. Dave, don't take our word for it, try it out on this applet and see for yourself. :) Posted by: Jason Malloy at July 11, 2003 04:05 AMWhat you say about doubling the probability makes sense, but is there an actual statistical study analagous to the situation described, that has proven it? Did Le'ts Make a Deal keep stats on it? Posted by: MaryClaire at July 11, 2003 07:03 AMDave B.ok I'm completely seriousi'll bet you $X by paypal right now. David B.  This might help. Suppose we have the original 3 doors and: 1) Monty always offers to let you change your first pick, but doesn't open any doors, and 2) you have a friend that operates a camera on the show. Your friend says "I know where the prize is and I'll help you. The director told me that when Monty offers to let you change your pick, I must point my camera at one of the two unselected doors. I'll always point to the one with the prize (unless your first pick was right, in which case I can't help you). You can see this in the monitor. Thus if the prize is behind *either* of the two doors you didn't pick at first, you're sure to win by changing your selection to the door in the monitor. The policy of always changing your pick to the one in the monitor increases your odds of winning from 1/3 to 2/3" Your say "great, but people might notice. Do what you said, but point to the other door, instead. I'll pick the one you don't point to. Same thing." And that is exactly what Monty is doing on the actual show when he opens one of the unselected doors, knowing it is empty. Posted by: BobJohnson at July 11, 2003 09:27 AM 

